Motion Graphs and Acceleration

A distance–time graph shows how far an object has travelled against the time taken. The shape of the line tells you how the object is moving.

The key rule is that the gradient (steepness) of a distance–time graph equals the speed of the object.

On a distance–time graph, speed = gradient. Divide the change in distance by the change in time over a section of the line.

Line on the graph	What the object is doing
Horizontal (flat)	stationary (speed = 0)
Straight, sloping	moving at constant speed
Steeper straight line	moving at a higher constant speed
Curved, getting steeper	speeding up (accelerating)

Worked example — a line rises from 0 m to 30 m between 0 s and 6 s. Find the speed.

$\text{speed}=\frac{\text{change in distance}}{\text{change in time}}=\frac{30}{6}=5\text{ m/s}$

The object moves at a constant 5 m/s.

(Higher Tier only) For a curved distance–time graph, the speed at a point is the gradient of the tangent drawn to the curve at that point.

Acceleration measures how quickly velocity changes. It is the change in velocity divided by the time taken.

$a=\frac{\Delta v}{t}$

where $a$ is acceleration in metres per second squared (m/s²), $\Delta v$ is the change in velocity in m/s, and $t$ is the time taken in seconds (s).

You must recall and apply this equation. $a=\Delta v/t$ is not given on the equation sheet.

A positive acceleration means speeding up; a negative acceleration (deceleration) means slowing down.

Worked example — a car speeds up from 8 m/s to 20 m/s in 4 s. Find its acceleration.

$a=\frac{\Delta v}{t}=\frac{20-8}{4}=\frac{12}{4}=3{\text{ m/s}}^2$

The car accelerates at 3 m/s².

Worked example — a train slows from 25 m/s to 10 m/s in 5 s.

$a=\frac{10-25}{5}=\frac{-15}{5}=-3{\text{ m/s}}^2$

The acceleration is −3 m/s²; the negative sign shows the train is decelerating.

A velocity–time graph shows how an object's velocity changes with time. It carries two pieces of information at once.

On a velocity–time graph, the gradient equals the acceleration. A steeper line means a greater acceleration; a horizontal line means constant velocity.

The graph below shows an object accelerating steadily, then travelling at constant velocity, then decelerating to rest.

Reading the first section: the velocity rises from 0 to 12 m/s in 4 s, so the acceleration is $12÷4=3$ m/s².

(Higher Tier only) The area under a velocity–time graph equals the distance travelled. Split the area into triangles and rectangles, or count squares for a curved line.

Worked example (HT) — find the total distance for the graph above by area.

• Rising triangle (0–4 s): $\frac{1}{2}\times 4\times 12=24$ m
• Middle rectangle (4–8 s): $4\times 12=48$ m
• Falling triangle (8–12 s): $\frac{1}{2}\times 4\times 12=24$ m

Total distance = $24+48+24=\text{96 m}$.

For an object with uniform (constant) acceleration, velocity, acceleration and distance are linked by a single equation.

$v^2-u^2=2as$

where $v$ is final velocity (m/s), $u$ is initial velocity (m/s), $a$ is acceleration (m/s²), and $s$ is distance (m).

This equation is given on the Physics equation sheet.

It is useful when you know the distance but not the time, so $a=\Delta v/t$ cannot be used directly.

Worked example — a car accelerates uniformly from rest ($u=0$) at 2 m/s² over 25 m. Find its final velocity.

$v^2-u^2=2as$ $v^2-0^2=2\times 2\times 25=100$ $v=\sqrt{100}=10\text{ m/s}$

The final velocity is 10 m/s.

Near the Earth's surface, objects in free fall (falling under gravity alone) accelerate at about 9.8 m/s². This value can be used as the acceleration $a$ in free-fall problems.

(Separate Physics only) Drawing and interpreting the velocity–time graph for terminal velocity, and explaining it in terms of forces, is AQA GCSE Physics only.

An object falling through a fluid (a liquid or gas, such as air) does not speed up forever. It reaches a steady maximum velocity called the terminal velocity.

An object reaches terminal velocity when the resultant force on it is zero.

When a skydiver first jumps, their weight is much larger than the small air resistance, so there is a large downward resultant force and they accelerate. As they speed up, air resistance (drag) increases. Eventually the upward drag grows until it equals the weight. The resultant force is now zero, so the acceleration is zero and the velocity stays constant at the terminal velocity.

At terminal velocity the two forces are equal and opposite, as shown above. On a velocity–time graph, the object accelerates (rising curve) and then levels off to a horizontal line once terminal velocity is reached. Opening a parachute increases the air resistance sharply, so the resultant force becomes upward, the skydiver decelerates, and a new, lower terminal velocity is reached.

1. Confusing distance–time and velocity–time graphs

On a distance–time graph the gradient is speed; on a velocity–time graph the gradient is acceleration and the area is distance. Check the axis labels before reading anything off.

2. Reading a flat velocity–time line as stationary

A horizontal line on a velocity–time graph means constant velocity, not zero velocity. Only a line sitting on the time axis (velocity = 0) means the object is at rest.

3. Forgetting to square-root in v² − u² = 2as

This equation gives you $v^2$, not $v$. Take the square root at the end to find the velocity itself.

4. Using v² − u² = 2as when acceleration is not uniform

This equation only works for constant (uniform) acceleration. If the acceleration changes, use graph areas or gradients instead.

5. Saying terminal velocity is when there is no force

At terminal velocity the resultant force is zero, but the weight and air resistance are both still acting; they are equal and opposite. The object moves at constant velocity, not because there are no forces, but because they balance.